这题其实就是反演一波就好了（那你还推了一下午+一晚上），不过第一次碰到\(O(n\log n)\)预处理分块和式的方法……
不知为啥我跟唐教主的题解推的式子不太一样……（虽然本质上可能是相同的吧）
那就写一写好了，顺便骗点访问量（逃
\[\begin{align}
\nonumber\text{Let}\space A=&\sum_{i=1}^n\sum_{j=1}^i i\sigma_1(ij),B=\sum_{i=1}^n i\sigma_1(i^2)\\
\nonumber\text{Then}\space Ans=&2A-B\\
\end{align}\]
\nonumber\text{Let}\space A=&\sum_{i=1}^n\sum_{j=1}^i i\sigma_1(ij),B=\sum_{i=1}^n i\sigma_1(i^2)\\
\nonumber\text{Then}\space Ans=&2A-B\\
\end{align}\]
\[\begin{align}
\nonumber A=&\sum_{i=1}^n\sum_{j=1}^i i\sigma_1(ij)\\
\nonumber =&\sum_{i=1}^n\sum_{j=1}^i i\sum_{p|i}\sum_{q|j}[(p,q)=1]\frac{pj}q\\
\nonumber =&\sum_{d=1}^n\mu(d)\sum_{i=1}^n\sum_{j=1}^i i\sum
_{p|i}\sum_{q|j}[d|(p,q)]\frac{pj}q\\
\nonumber =&\sum_{d=1}^n\mu(d)\sum_{d|p}p\left(\sum_{p|i}i\right)\sum_{d|q}\sum_{q|j}^i \frac j q\\
\nonumber =&\sum_{d=1}^n\mu(d)\sum_{p=1}^{\left\lfloor\frac n d\right\rfloor}dp\left(\sum_{p|i}^{\left\lfloor\frac n d\right\rfloor}di\right)\sum_{q=1}^{\left\lfloor\frac n d\right\rfloor}\sum_{q|j}^i \frac{dj}{dq}\\
\nonumber =&\sum_{d=1}^n\mu(d)\sum_{p=1}^{\left\lfloor\frac n d\right\rfloor}dp\left(\sum_{p|i}^{\left\lfloor\frac n d\right\rfloor}di\right)\sum_{q=1}^i\sum_{q|j}^i \frac j q\\
\nonumber =&\sum_{d=1}^n\mu(d)d^2\sum_{i=1}^{\left\lfloor\frac n d\right\rfloor}i\left(\sum_{p|i}p\right)\sum_{j=1}^i\sum_{q|j}\frac j q\\
\nonumber =&\sum_{d=1}^n\mu(d)d^2\sum_{i=1}^{\left\lfloor\frac n d\right\rfloor}i\sigma_1(i)\sum_{j=1}^i\sigma_1(j)\\
\nonumber =&\sum_{d=1}^n\mu(d)d^2\sum_{i=1}^{\left\lfloor\frac n d\right\rfloor}i\sigma_1(i)S_{\sigma_1}(i)
\end{align}\]
\nonumber A=&\sum_{i=1}^n\sum_{j=1}^i i\sigma_1(ij)\\
\nonumber =&\sum_{i=1}^n\sum_{j=1}^i i\sum_{p|i}\sum_{q|j}[(p,q)=1]\frac{pj}q\\
\nonumber =&\sum_{d=1}^n\mu(d)\sum_{i=1}^n\sum_{j=1}^i i\sum
_{p|i}\sum_{q|j}[d|(p,q)]\frac{pj}q\\
\nonumber =&\sum_{d=1}^n\mu(d)\sum_{d|p}p\left(\sum_{p|i}i\right)\sum_{d|q}\sum_{q|j}^i \frac j q\\
\nonumber =&\sum_{d=1}^n\mu(d)\sum_{p=1}^{\left\lfloor\frac n d\right\rfloor}dp\left(\sum_{p|i}^{\left\lfloor\frac n d\right\rfloor}di\right)\sum_{q=1}^{\left\lfloor\frac n d\right\rfloor}\sum_{q|j}^i \frac{dj}{dq}\\
\nonumber =&\sum_{d=1}^n\mu(d)\sum_{p=1}^{\left\lfloor\frac n d\right\rfloor}dp\left(\sum_{p|i}^{\left\lfloor\frac n d\right\rfloor}di\right)\sum_{q=1}^i\sum_{q|j}^i \frac j q\\
\nonumber =&\sum_{d=1}^n\mu(d)d^2\sum_{i=1}^{\left\lfloor\frac n d\right\rfloor}i\left(\sum_{p|i}p\right)\sum_{j=1}^i\sum_{q|j}\frac j q\\
\nonumber =&\sum_{d=1}^n\mu(d)d^2\sum_{i=1}^{\left\lfloor\frac n d\right\rfloor}i\sigma_1(i)\sum_{j=1}^i\sigma_1(j)\\
\nonumber =&\sum_{d=1}^n\mu(d)d^2\sum_{i=1}^{\left\lfloor\frac n d\right\rfloor}i\sigma_1(i)S_{\sigma_1}(i)
\end{align}\]
\[\begin{align}
\nonumber B=&\sum_{i=1}^n i\sigma_1(i^2)\\
\nonumber =&\sum_{i=1}^n i\sum_{p|i}\sum_{q|j}[(p,q)=1]\frac{pi}q\\
\nonumber =&\sum_{d=1}^n\mu(d)\sum_{i=1}^n i\sum_{p|i}\sum_{q|j}[d|(p,q)]\frac{pi}q\\
\nonumber =&\sum_{d=1}^n\mu(d)\sum_{d|i} i\sum_{d|p|i}\sum_{d|q|j}\frac{pi}q\\
\nonumber =&\sum_{d=1}^n\mu(d)d^2\sum_{i=1}^{\left\lfloor\frac n d\right\rfloor} i\sum_{p|i}\sum_{q|j}p\frac i q\\
\nonumber =&\sum_{d=1}^n\mu(d)d^2\sum_{i=1}^{\left\lfloor\frac n d\right\rfloor} i\sigma_1^2(i)\\
\end{align}\]
\nonumber B=&\sum_{i=1}^n i\sigma_1(i^2)\\
\nonumber =&\sum_{i=1}^n i\sum_{p|i}\sum_{q|j}[(p,q)=1]\frac{pi}q\\
\nonumber =&\sum_{d=1}^n\mu(d)\sum_{i=1}^n i\sum_{p|i}\sum_{q|j}[d|(p,q)]\frac{pi}q\\
\nonumber =&\sum_{d=1}^n\mu(d)\sum_{d|i} i\sum_{d|p|i}\sum_{d|q|j}\frac{pi}q\\
\nonumber =&\sum_{d=1}^n\mu(d)d^2\sum_{i=1}^{\left\lfloor\frac n d\right\rfloor} i\sum_{p|i}\sum_{q|j}p\frac i q\\
\nonumber =&\sum_{d=1}^n\mu(d)d^2\sum_{i=1}^{\left\lfloor\frac n d\right\rfloor} i\sigma_1^2(i)\\
\end{align}\]
\[Ans=2A-B\\=\sum_{d=1}^n\mu(d)d^2\sum_{i=1}^{\left\lfloor\frac n d\right\rfloor}i\sigma_1(i)\left(2S_{\sigma_1}(i)-\sigma_1(i)\right)
\]
\]
到了这里就可以用\(O(n)\)预处理+单次询问\(O(\sqrt n)\)的经典做法了，总复杂度\(O(n+T\sqrt n)\)。
但是这样还是有点慢……常数优化到一定程度之后极限数据仍然要跑2s+（卡常技巧不行……逃），看来常数优化玩不了了，只能试试别的做法。
考虑枚举\(d\)和\(k=\left\lfloor\frac n d\right\rfloor\)，并考虑它们能对哪些\(n\)作出贡献。不难发现，如果\(\left\lfloor\frac n d\right\rfloor=k\)，那么一定有\(n\in[dk,d(k+1))\)，再稍加观察就能得到每对\((d,k)\)都会对\(\ge dk\)的所有\(n\)作出贡献，因此枚举所有\(d,k\)并差分一下，最后求一遍前缀和即可。
这样就能做到预处理\(O(n\log n)\)，询问\(O(1)\)了，不用卡常也可以过。
#pragma GCC optimize("Ofast")
#include<stdio.h>
#define int unsigned
using namespace std;
template<class T>inline void readint(T &v){
int x=0;
char c=getchar();
while(c<48)c=getchar();
while(c>47){
x=x*10+(c^48);
c=getchar();
}
v=x;
}
template<class T>inline void writeint(T x){
static int a[25];
if(!x)putchar('0');
else{
int i=0;
while(x){
a[++i]=x%10;
x/=10;
}
while(i)putchar(a[i--]^48);
}
}
const int maxn=1000010,p=1000000007;
void get_table(int);
bool notp[maxn];
int prime[maxn],mu[maxn],sigma[maxn],ans[maxn],f[maxn];
signed main(){
get_table(1000000);
int T;
readint(T);
for(int k=1;k<=T;k++){
int n;
readint(n);
putchar('C');
putchar('a');
putchar('s');
putchar('e');
putchar(' ');
putchar('#');
writeint(k);
putchar(':');
putchar(' ');
writeint(ans[n]);
putchar('\n');
}
return 0;
}
void get_table(int n){
mu[1]=sigma[1]=1;
for(int i=2;i<=n;i++){
if(!notp[i]){
prime[++prime[0]]=i;
mu[i]=-1;
sigma[i]=i+1;
f[i]=1;
}
for(int j=1;j<=prime[0]&&i*prime[j]<=n;j++){
notp[i*prime[j]]=true;
if(i%prime[j]){
mu[i*prime[j]]=-mu[i];
sigma[i*prime[j]]=(long long)sigma[i]*(prime[j]+1)%p;
f[i*prime[j]]=i;
}
else{
if(f[i]==1)sigma[i*prime[j]]=((long long)prime[j]*sigma[i]+1)%p;
else sigma[i*prime[j]]=(long long)sigma[i/f[i]*prime[j]]*sigma[f[i]]%p;
f[i*prime[j]]=f[i];
}
}
}
for(int i=2,sum=1;i<=n;i++){
sum+=sigma[i];
if(sum>=p)sum-=p;
sigma[i]=(long long)i*sigma[i]%p*(2ll*sum-sigma[i]+p)%p;
}
for(int i=1;i<=n;i++){
int t=((long long)i*i%p*(signed)mu[i]+p)%p;
for(int j=1;i*j<=n;j++)ans[i*j]=(ans[i*j]+(long long)t*sigma[j])%p;
}
for(int i=2;i<=n;i++){
ans[i]+=ans[i-1];
if(ans[i]>=p)ans[i]%=p;
}
}